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sergeinik [125]
3 years ago
13

The net potential energy EN between two adjacent ions, is sometimes represented by the expression

Engineering
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

as answered in the attached file.

Explanation:

The detailed steps, derivation and appropriate differentiation is as shown in the attachment

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The second programming project involves writing a program that accepts an arithmetic expression of unsigned integers in postfix
Tpy6a [65]

Answer:

Explanation:

Note: In case of any queries, just comment in box I would be very happy to assist all your queries

SourceCode:

// MyGUI.java:

// Import packages

import java.awt.FlowLayout;

import java.awt.GridLayout;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.EmptyStackException;

import java.util.Stack;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JTextField;

import javax.swing.SwingConstants;

// Declaare and define the class MyGUI

abstract class MyGUI extends JFrame implements ActionListener {

JTextField userInput;

JLabel inputDescLbl, resultLbl;

JPanel inputPanel, resultPanel;

JButton evlBtn;

Stack<Object> stk;

// Define the constructor MyGUI

MyGUI() {

super("Tree Address Generator");

inputPanel = new JPanel(new FlowLayout());

resultPanel = new JPanel(new FlowLayout());

setLayout(new GridLayout(2, 1));

userInput = new JTextField(20);

inputDescLbl = new JLabel("Enter Postfix Expression:");

evlBtn = new JButton("Construct Tree");

evlBtn.addActionListener(this);

resultLbl = new JLabel("Infix Expression:", SwingConstants.LEFT);

add(inputPanel);

add(resultPanel);

inputPanel.add(inputDescLbl);

inputPanel.add(userInput);

inputPanel.add(evlBtn);

resultPanel.add(resultLbl);

stk = new Stack<Object>();

}

}

//Stack.java:

// Declare and define the class Stack

class Stack {

private int[] a;

private int top, m;

public Stack(int max) {

m = max;

a = new int[m];

top = -1; }

public void push(int key) {

a[++top] = key; }

public int pop() {

return (a[top--]); }

}

// Declare and define the class Evaluation()

class Evaluation {

public int calculate(String s) {

int n, r = 0;

n = s.length();

Stack a = new Stack(n);

for (int i = 0; i < n; i++) {

char ch = s.charAt(i);

if (ch >= '0' && ch <= '9')

a.push((int) (ch - '0'));

else if (ch == ' ')

continue;

else {

int x = a.pop();

int y = a.pop();

switch (ch) {

case '+':

r = x + y;

break;

case '-':

r = y - x;

break;

case '*':

r = x * y;

break;

case '/':

r = y / x;

break;

default:

r = 0;

}

a.push(r);

}

}

r = a.pop();

return (r);

}

}

// PostfixToInfix.java:

// Import packages

import java.util.Scanner;

import java.util.Stack;

// Declare and define the class PostfixToInfix

class PostfixToInfix {

// Determine whether the character entered is an operator or not

private boolean isOperator(char c) {

if (c == '+' || c == '-' || c == '*' || c == '/' || c == '^')

return true;

return false;

}

// Declare and define the convert()

public String convert(String postfix) {

Stack<String> s = new Stack<>();

for (int i = 0; i < postfix.length(); i++) {

char c = postfix.charAt(i);

if (isOperator(c)) {

String b = s.pop();

String a = s.pop();

s.push("(" + a + c + b + ")");

} else

s.push("" + c);

}

return s.pop();

}

// Program starts from main()

public static void main(String[] args) {

PostfixToInfix obj = new PostfixToInfix();

Scanner sc = new Scanner(System.in);

// Prompt the user to enter the postfix expression

System.out.print("Postfix : ");

String postfix = sc.next();

// Display the expression in infix expression

System.out.println("Infix : " + obj.convert(postfix));

}

}

Output:

e Console X terminated PostfixTolnfix [Java Application] C:\Program Files\Java\jrel.8.0_121\bin\javaw.exe Postfix : ABD++C-D/ .

3 0
3 years ago
Which of the following would NOT fall into one of the WHMIS 1988 classes?
sweet [91]
A. Compressed Gas Yup
8 0
3 years ago
Read 2 more answers
The bulk modulus of a fluid if it undergoes a 1% change in volume when subjected to a pressure change of 10,000 psi is (a) 0.01
Veseljchak [2.6K]

Answer:

The required bulk modulus is 10^{6} Psi. So, the answer is non of these.

Explanation:

Change in pressure of the fluid is directly proportional to the volumetric strain. The constant of proportionality is the bulk modulus of the fluid.

Step1

Given:

Percentage change in volume is 1%.

Change in pressure is 10000 Psi.

Calculation:

Step2

Volumetric strain is calculated as follows:

\frac{\bigtriangleup V}{V}=\frac{1}{100}

\frac{\bigtriangleup V}{V}=0.01

Step3

Bulk modulus is calculated as follows:

\frac{\bigtriangleup V}{V}=0.01

\frac{\bigtriangleup V}{V}=0.01

10000=K\times0.01

K = 1000000 Psi.

Thus, the required bulk modulus is 10^{6} Psi.

3 0
3 years ago
One reason that driving lends itself to aggressive behavior is:
Maurinko [17]

Answer:

C. Drivers feel empowered by anonymity

Explanation:

Sorry for the late answer hope it helps you and others who need it.

STay Safe and healthy!

5 0
3 years ago
Read 2 more answers
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected
Pavlova-9 [17]

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d)  = ?

a) for stress

The stress equation is given by:

S = \frac{P}{A}

A is the area = πd²/4 = (3.14 × d²)/4

S = \frac{P}{(\frac{3.14*d^{2} }{4}) }

S = \frac{4P}{{3.14*d^{2} } }

3.14*S*{d^{2}} = {4P}

{d^{2}} =\frac{4P}{3.14*S}

d=  \sqrt{\frac{4P}{3.14*S} }

Substituting the values, we get

d=  \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }

d=  \sqrt{\frac{12000 }{125600000  } }

d=  \sqrt{9.55*10^{-5}  }

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

3 0
3 years ago
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